Part (a)
There are 7 red out of 7+3 = 10 total
<h3>Answer: 7/10</h3>
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Part (b)
We have 3 green out of 10 total
<h3>Answer: 3/10</h3>
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Part (c)
3/10 is the probability of getting green on any selection. This is because we put the first selection back (or it is replaced with an identical copy)
So (3/10)*(3/10) = 9/100 is the probability of getting two green in a row.
<h3>Answer: 9/100</h3>
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Part (d)
Similar to part (c) we have 7/10 as the probability of getting red on each independent selection.
(7/10)*(7/10) = 49/100
<h3>Answer: 49/100</h3>
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Part (e)
7/10 is the probability of getting red and 3/10 is the probability of getting green. Each selection is independent of any others.
(7/10)*(3/10) = 21/100
<h3>Answer: 21/100</h3>
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Part (f)
We have the exact same set up as part (e). Notice how (7/10)*(3/10) is the same as (3/10)*(7/10).
<h3>Answer: 21/100</h3>
I would like to ask you to go to the vet, or attempt to, as this might be a worm in the eye. The same thing happened to my dog.
Answer:
0.28cm/min
Step-by-step explanation:
Given the horizontal trough whose ends are isosceles trapezoid
Volume of the Trough =Base Area X Height
=Area of the Trapezoid X Height of the Trough (H)
The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)
The Volume of water in the trough at any time
=8h(8+2x)
V=64h+16hx
We are not given a value for x, however we can express x in terms of h from Figure 3 using Similar Triangles
x/h=1/4
4x=h
x=h/4
Substituting x=h/4 into the Volume, V
h=3m,
dV/dt=25cm/min=0.25 m/min
=0.002841m/min =0.28cm/min
The rate is the water being drawn from the trough is 0.28cm/min.
75° because they are alternate interior angles
